Acopper refinery produces a copper ingot weighing 150 lb. if the copper is drawn into wire whose diameter is 8.25mm, how many feet of copper can be obtained from the ingot? the density of copper is 8.94 g/cm^3. (assume that the wire is a cylinder whose volume is v=3.14r^2h, where r is its radius and h is its height or length.

 you have V = π x r² x L right? 
and you know mass and density and density = mass / volumeright? 
rearranging...(ρ = density, m = mass, V = volume) 
V = m / ρ V = π x r² x L 
m / ρ = π x r² x L 
but r = D/2 {where D is diameter} so r² = (D/2)² = D² / 4 
m / ρ = π x (D² / 4) x L 
L = 4 x m / (π x ρ x D²) 
L = 4 x m x (1/ π) x (1/ρ) x (1/ D²) {makes it easier to see next step} 
L = 4 x (150 lbs) x (454 g / lb) x ( 1/ 3.1415) x (1 cm³ / 8.94 g) x (1 / 8.25 mm)² x (10 mm / 1 cm)² = 14250 cm 
still with me? 
14250 cm x (1 in / 2.54 cm) x (1 ft / 12 in) = 468 ft{ 3 sig figs} 
alternately, you can break up the steps.. 
150 lbs x (454 g/lb) = 68,100 g Cu 
68,100g Cu x (1 cm³ / 8.94 g) = 7617 cm³ copper 
diameter of wire = 8.25 mm = .825 cm so radius of wire = 0.4125 cm 
7617 cm³ = (3.14) x r² x L = 3.14 x (.4125 cm)² x L 
7617 cm³ = 0.534 cm² L 
L = 7617 cm³ / 0.534 cm² = 14250 cm 
14250 cm x (1 in / 2.54 cm) x (1 ft / 12 in) = 468 ft{ 3 sig figs}